Self-Assessment Quiz: Why Quantum Error Correction Is Necessary
Twenty questions on decoherence, error channels, no-cloning, and why quantum errors are harder than classical ones. Aim for 16+.
Question 1
$T_1$ characterizes:
A) Energy relaxation ($|1\rangle \to |0\rangle$) B) Dephasing only C) Gate time D) Readout duration
Question 2
$T_2$ characterizes:
A) Loss of phase coherence B) Energy relaxation only C) Measurement error D) Gate fidelity
Question 3
The relationship between them is:
A) $T_2 \le 2T_1$ B) $T_2 \ge 2T_1$ C) $T_2 = T_1$ D) Unrelated
Question 4
A qubit idling for time $t$ retains coherence approximately:
A) $e^{-t/T_2}$ B) $1 - t/T_2$ C) $t/T_2$ D) Constant
Question 5
Quantum errors are continuous, unlike classical bit flips. Error correction handles this by:
A) Measuring the syndrome, which discretizes the error into a Pauli B) Ignoring small errors C) Rounding amplitudes D) It cannot
Question 6
The three basic single-qubit Pauli errors are:
A) $X$ (bit flip), $Z$ (phase flip), $Y$ (both) B) $H$, $S$, $T$ C) CNOT, SWAP, Toffoli D) Measurement, reset, idle
Question 7
Correcting arbitrary single-qubit errors requires correcting:
A) Only $X$ B) Both $X$ and $Z$ (which implies $Y$) C) Only $Z$ D) All infinitely many rotations individually
Question 8
The no-cloning theorem prevents:
A) Simple repetition-code redundancy by copying an unknown state B) All error correction C) Measurement D) Entanglement
Question 9
QEC works around no-cloning by:
A) Copying the state B) Spreading information across an entangled block without copying it C) Measuring the state D) Using classical bits
Question 10
Syndrome measurement must:
A) Reveal which error occurred without revealing the encoded logical state B) Measure the logical qubit directly C) Destroy the encoding D) Be classical
Question 11
If syndrome measurement revealed the logical state:
A) Nothing would change B) It would collapse the superposition, destroying the computation C) Correction would be easier D) The code would be more efficient
Question 12
Amplitude damping describes:
A) Energy loss to the environment B) Phase noise only C) Readout error D) Gate over-rotation
Question 13
Coherent errors (e.g. systematic over-rotation) differ from stochastic errors in that they:
A) Accumulate linearly in amplitude, so can add quadratically in probability — worse than random errors B) Are always smaller C) Cannot be corrected D) Do not affect fidelity
Question 14
Crosstalk is:
A) Unwanted interaction between qubits during gates or idling B) Classical communication C) A readout technique D) A type of measurement
Question 15
The threshold theorem states that:
A) Arbitrary-length computation is possible if physical error is below a threshold, with polylogarithmic overhead B) Errors cannot be corrected C) Error correction requires no overhead D) The threshold is zero
Question 16
True or false: Quantum error correction can correct errors without knowing the encoded state.
Question 17
True or false: Measuring the syndrome collapses the logical qubit.
Question 18
True or false: Error correction can be applied once at the end of a computation.
Question 19
Short answer. Explain why continuous quantum errors can be corrected by a code designed for discrete Pauli errors.
Question 20
Short answer. Why are coherent errors more dangerous than incoherent errors of the same magnitude?
Answer Key
| Q | Ans | Note |
|---|---|---|
| 1 | A | Longitudinal relaxation toward the ground state. |
| 2 | A | Transverse decoherence — loss of the relative phase that carries superposition. |
| 3 | A | $T_2 \le 2T_1$: relaxation necessarily destroys phase, and pure dephasing shortens $T_2$ further. |
| 4 | A | Exponential decay of the off-diagonal density-matrix elements. |
| 5 | A | The syndrome projects a continuous error onto a discrete Pauli — the key insight making QEC possible at all. |
| 6 | A | $Y = iXZ$, so $X$ and $Z$ generate everything. |
| 7 | B | Correcting the generators corrects any linear combination — including arbitrary small rotations. |
| 8 | A | You cannot make backup copies of an unknown state, so the classical repetition strategy is unavailable. |
| 9 | B | Information is delocalized across the block: no single physical qubit holds a copy, and the logical state lives in correlations. |
| 10 | A | The defining requirement of a stabilizer measurement. |
| 11 | B | Collapse would destroy exactly the superposition the computation depends on. |
| 12 | A | The $T_1$ channel. |
| 13 | A | Amplitudes add coherently, so $N$ over-rotations of $\epsilon$ give error $\sim(N\epsilon)^2$ rather than $N\epsilon^2$. |
| 14 | A | A significant and often under-modelled error source in dense superconducting layouts. |
| 15 | A | The foundational result: fault tolerance is possible in principle, given a good enough physical error rate. |
| 16 | True | Syndromes identify the error while leaving the logical information untouched — precisely what makes QEC work. |
| 17 | False | It collapses the error onto a Pauli while leaving the logical state coherent. Stabilizer measurements commute with logical operators by construction. |
| 18 | False | Errors accumulate continuously and would exceed the code's correction capacity. Syndrome extraction runs repeatedly throughout — thousands of rounds per logical operation. |
| 19 | — | An arbitrary single-qubit error can be expanded in the Pauli basis: $E = \alpha I + \beta X + \gamma Y + \delta Z$. Applying $E$ to an encoded state gives a superposition of the four error branches. Measuring the syndrome projects onto one branch, collapsing the continuous error to a discrete Pauli with probability given by its amplitude — which the code then corrects. Linearity plus measurement discretizes the continuum, so a code correcting $X$ and $Z$ corrects everything. |
| 20 | — | Stochastic errors add in probability: $N$ errors of probability $p$ give total error $\approx Np$. Coherent errors add in amplitude: $N$ systematic over-rotations of angle $\epsilon$ produce a total rotation $N\epsilon$, so error probability $\approx (N\epsilon)^2 = N^2\epsilon^2$ — quadratically worse in $N$. Coherent errors also evade some benchmarking (they partially cancel in symmetric sequences), so they are both more damaging and harder to detect. Randomized compiling deliberately converts coherent errors into stochastic ones, accepting a worse average to gain predictable scaling. |